We must have $n$ divide it for $a=2$: $$2^{25}-2=2(2^{12}-1)(2^{12}+1)=2(2^6-1)(2^6+1)(2^4+1)(2^8-2^4+1);$$$$2^{25}-2=2\cdot 3^2 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 241.$$Letting $a$ be a primitive root mod $17$ and $241$ will result in a number of the form $a^{25}-a$ which is multiple of none of these primes. Also, writing $a(a^{24}-1)$, we see that $24$ is multiple of $\phi(p)$ for $p \in \{2,3,5,7,13\}$, so $a^{25}-a$ is divisible by all these primes, for every $a$. However, it is not always divisible by $3^2$ (take $a=3$, for example). Therefore, $2\cdot 3\cdot 5 \cdot 7 \cdot 13 =2730$, which has $\boxed{31}$ divisors greater than $1$, is the largest number guaranteed to always divide $a^{25}-a$.